Show that $z - xy$ is in the ideal $I = \langle 2x-1, 3y-1, 6z-1 \rangle$ in $\mathbb{Z}[x,y,z]$

Question

This is inspired by an exercise in Eisenbud's text on commutative algebra. In the text, he gives the following alternative characterization of localization: for any commutative ring $R$ with unity and any subset $U$ of $R$ (not necessarily multiplicatively closed as the text defines $R[U^{-1}] = R[\bar{U}^{-1}]$), there is a natural isomorphism

\begin{align} R[U^{-1}] \cong R[\{x_{u\}_{u \in U}}]/\langle\{ux_{u}-1\}_{u \in U}\rangle. \end{align}

Applying this to $R = \mathbb{Z}$ and $U = \{2,3,6\}$, this isomorphism becomes

\begin{align} \mathbb{Z}[U^{-1}] \cong \mathbb{Z}[x,y,z]/\langle 2x-1, 3y-1, 6z-1\rangle. \end{align}

Intuitively, $x = 1/2$, $y = 1/3$, and $z = 1/6$. It seems natural, then, that $z-xy$ should be in the ideal $I = \langle 2x-1, 3y-1, 6z-1\rangle$ of $\mathbb{Z}[x,y,z]$. I used Sage to verify this, but I can't think of a way to express $z-xy$ in terms of the generators of $I$ (note carefully, that I'm considering $I$ as an ideal of $\mathbb{Z}[x,y,z]$)

Answer 1

\begin{align} z-xy&=z-6xyz+xy(6z-1)\\ &=z-(2x)(3y)z+xy(6z-1)\\ &=z-3(2x-1)yz-3yz+xy(6z-1)\\ &=-(3y-1)z-3(2x-1)yz+xy(6z-1) \end{align}

Answer 2

Note that $(6z-1)-2x(3y-1)-(2x-1)=6z-6xy$ is in the ideal

Now using the intuition $z=\frac 16$ we multiply by $z$ to find also $6z^2-6xyz$ in the ideal and this is $z(6z-1)-xy(6z-1)+z-xy$

You should be able to work backwards to the expression you need.